Group
Comparing speeds on our roadster with and without rear fenders and no other
changes the computed Cd of the rear tires is near the bottom of the quoted
range. That number seems low, but that's what the computation shows.
Dave the Hayseed
----- Original Message -----
From: "Skip Higginbotham" <saltrat@pro-blend.com>
To: "DrMayf" <drmayf@teknett.com>; <land-speed@autox.team.net>
Sent: Monday, March 04, 2002 6:38 PM
Subject: Re: Tires
> Mayf,
> Congratulations on your home and shop!!! Having done that a couple of
> times, I know how much fun it is. (-:(-: Would love to see it later on
this
> month when we are there.
>
> Thank you for taking your time to take a stab at this problem. From your
> analysis and conjectural deduction, it would seem that covering/enclosing
> the upper half is the most important as it reduces the drag by about ?%
> (may be about 65% or so) because the free stream air now only interferes
> with the enclosure and not the rotating tire. The rotating tire and the
> shear between it and the enclosure have a drag loss but the fenderwell
> (enclosure) vents can reduce this some. We will add the vents like the
> sports cars do. How far down the tire should the enclosure go depends on,
I
> suppose, the diameter of the tire.....or does it? As the tire surface
> airspeed approaches 0 (when the tire is contacting the surface, no
> slippage) drag also approaches 0. So the lower part (?%) can run in free
> stream air with little drag penalty.
> We are building enclosures for the rear tires on the "Rose" and results of
> this discussion will directly effect the amount of tire that gets
enclosed.
> Right now I would cover 75% of the tire and leave the bottom 7-8" out in
> the open. Simplifies the fairing problem a lot! I'll see if my little mind
> can figure out the differential airspeed for the lower 8".
>
> How any of this would work in a fenderwell that is out of free stream air
> absolutely baffles me. But it might help some if the airflow in the
> fenderwell is significant and the direction can be established. Prolly
need
> some tunnel time for this one.
>
> I have seen Cd for open tires that vary from .19 to .7.......anybody have
> some measured data?
>
> Skip (This is what makes this sport fun!)
>
>
>
>
> At 02:53 PM 3/4/02 -0800, you wrote:
> >Skip, you asked some really great questions in this post! I am not sure I
> >can answer them in other than general terms and thoughts. See the
thoughts
> >interspersed with what you wrote below...
> > Subject: RE: Tires
> >
> >
> >> Hello Ed,
> >> Tire frontal area is a really big issue at any speed as drag increases
> >> greatly with increase in speed and the Cd is relatively high.......but
how
> >> about considering this......the top of the tire is moving twice as fast
as
> >> the car, relative to the wind, and the bottom of the tire is not moving
at
> >> all relative to the wind (no slippage). So if you cover the top of the
> >> tire, most of the skin drag would be eliminated? Here's one to
calculate,
> >> Mayf. The desired answer might be "How much of the tire to cover?" or
what
> >> is the real drag of an open tire?
> >
> >IMHO, the drag on a tires is complex to say the least. Maybe that is why
> >there is so little data available on the subject. The tire's drag
> >coefficient must be somwhat like sticking a flat plate into the wind and
> >those have high drag numbers. Add in rotational flow due to the boundary
> >layer on the tire and it is probably not solveable on your calculator. I
had
> >a problem like this on a project once and it involved circular flow
patterns
> >and circular shock waves. It was the inverse of this problem and it
required
> >3D navier stokes equations and lots of computing power to solve. I know
some
> >out there have raised an eyebrow about the top of the tire speed you
> >mentioned so let's make sure everyone is confident that this is true.
First,
> >the center of the wheel/tire is moving at the same speed as the car. Then
> >you add the component due to rotation of the tire which is also the same
as
> >the forward speed (if someone needs this off line, let me know and I'll
give
> >your the math analysis that proves it). So the "relative wind" at the top
of
> >a tire is 2 times the forward speed of the car. So how much "fender" to
add?
> >This is a gut feel on my part but I would if the rules allowed it and
there
> >was room in the fenderwell I would start at the 3:00 o'clock position
cover
> >the top and front down to about the 8:00 o'clock position. This would
leave
> >a about 150 degrees of the full wheel open. The back would be open to let
> >that pesky salt fly off of the tire and the front being closed would IMHO
> >help to reduce the circular air flow from mixing with the fenderwell
flow.
> >This would cause a LOT of turbulence and turbulence is drag. A bigger
> >question is how close to the tire to make the fender? Well, remember that
> >the air attached to the tire is going like crazy and the air on the inner
> >side of the fender is stopped. Like the flow under your car. So there is
a
> >high shear in this air and that also causes drag. So what's a mother to
do?
> >Well, ever watched the world sports car classes? They have full fenders
but
> >the fender tops have rear facing louvers over the top side of the tire. I
> >thinks this is to reduce the turbulent air inside the fender. The air
flows
> >up and over the top of the fender and creates a low pressure at that
point
> >and sucks out the air inside the fender leaving a low pressure low drag
> >condition inside the fender. Make sense? We are talking computational
fluid
> >dynamics here and that button isn't on my calculator so I cannot solve...
> >But that is what I would do. Even on a full body car.
> >
> >> The top of our tires were going 542 MPH at the "5". The top of Law's
tires
> >> were going 700 MPH. Hammond's at 600. I wonder what sonic shock waves
will
> >> do to tire tread...or.....how narrow a tire can we use and still be
safe
> >> from excessive distortion? Seems we had a thread about flexing some
months
> >> ago......wonder what happened? I seem to remember some sensitivity
about
> >> it.....
> >
> >Ok, one thing we again need to do is make sure everyone is on the same
page
> >when we talk about supersonic conditions. The speed of sound is dependent
> >solely on the temperature of the air. Skeptics may write off line and
I'll
> >send the math... But it breaks down to this:
> >a = 49.018 (sqrt of 459.6 + T). So on a day when the temperature is 95
> >degrees F then the speed of sound is 49.018 * sqrt (554.6) or a = 1154.4
> >ft/second. A side thought here...you'd think that it would be better to
run
> >for the absolute LSR record (Thrust) when the temps were down because the
> >sonic velocity is down also. But remember that as temp goes down, the air
> >density goes up and hence drag goes up...it is a real trade off. Now
that
> >we understand the definition of the speed of sound, MAch number is the
> >vehicle speed divided by the speed of sound. That's why when Thrust went
> >fast, they had to wait a second or two while the speed of sound for that
run
> >was calculated so they could figure the Mach number... So how fast is
1154
> >ft/second? Well, it is 787 mph. So the toip of a tire on a 95 degree day
> >would have to be travelling 787 mph to get to Mach 1 and have a standing
> >normal shock wave on the top of the tire. But it is a very short shock
wave
> >because it is only the air that is attached to the tire's boundary layer
> >that is going that fast. Just a few millimeters away from the tire, the
air
> >is slowing because it is standing still either by virtue of the fender"
or
> >free air stream. Now remember, this is all conjecture on my part! YMMV.
> >
> >As to tires, well, I would be more concerned with the centrifugal force
on
> >the tread section. Here is why: F = mass * tire radius * omega^2 (omega
is 2
> >* pi * revs per sec of tire). As an example, lets say the tire is 28
inches
> >in diameter and weighs what? 24 pounds? Lets assume that the sidewalls of
> >the tire weigh 1/3 of the total weight, so the "tread section" then
weighs
> >16 pounds. Lets pretend that this tire tread is 8 inches wide. At 28
inches
> >in diameter, the circumference of the tire is 2 * pi * 14 = 88 inches. So
a
> >one inch wide strip of tread weighs about 0.18 pounds. Or, since our
tread
> >is 8 inches wide, then one square inch of tread would weigh 0.023 pounds.
So
> >now all we have to do is figure the rotational speed of the tire to
figure
> >out how much centrifugal force there is on that 1 square inch of tread.
> >Let's assume 300 mph or 440 ft/sec. Divide this by the circumference of
the
> >tire in feet and get 440 / 7.33 = 60 revolutions per second! Same as the
> >frequency of the AC power in your house. Pretty fast. Moving on, F =
> >0.023/32.2 * 1.16 ft * (2*pi* 60)^2
> >
> >F = 0.023/32.2 * 1.16 * (376.99 * 376.99) = 117.75 pounds! on that little
> >one square inch! ( the 0.023 was the weight of the section, 32.2 converts
it
> >to mass units, 1.16 is the tire radius in feet, pi = 3.1416, 60 was the
> >revolutions per second of the tire).
> >
> >Since I used a one square inch of tread, this is the same as saying that
the
> >tire is pressurized internally to nearly 120 psig when going 300 mph! And
> >that is on top of what you started with! So start with 60 psig then you
have
> >the equivalent of 180 psig acting on your tires! The tread is under a
> >tremendous strain. Because most tires are radial (are the Goodyears? I
think
> >so), the sidewall cord keepd the edge of the tread from expanding as much
as
> >the center of the tread so the tire takes on an oval shape. Reducing the
> >contact patch and reducing rolling resistance a whole hell of a lot!
IMHO.
> >And if you are adventurous, you can make a stab at how the salt stays in
a
> >treaded tire at these speeds...I don't think it does, I think it comes
off
> >nearly as soon as it sticks on the tire...
> >
> >Was this mental masturbation enough for everyone? Fun for me, was it good
> >for you?
> >
> >mayf, the red necked ignorant desert rat in Pahrump where the new house
> >power is on, water runs from the well, counter tops, sinks, and
associated
> >hardware is in, base boards are going in, the shop under sink hot water
> >heater is in (main house has tankless water heaters), toilets are all in
and
> >have water in them, getting closer....Maybe by the end of next week?
> >nah...bet it is 2 more weeks...
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